Question: $\sum\limits_{k=0}^{{39}}{{{8\left(-\dfrac78\right)^{k}}}} \approx$ Choose 1 answer: Choose 1 answer: (Choice A) A $0.24$ (Choice B) B $ 4.25 $ (Choice C) C $4.29$ (Choice D) D $64.31$
What is the question asking for? The question is asking for the sum of the values of $8\left(-\dfrac78\right)^k$ from $k = 0$ to $k = 39$ : $8\left(-\dfrac78\right)^0 + 8\left(-\dfrac78\right)^1 +... +8\left(-\dfrac78\right)^{39} $ The series is geometric because the formula $8\left(-\dfrac78\right)^k$ is an exponential function of $k$. Formula for geometric series The sum $S_n$ of a finite geometric series is $S_n = \dfrac{a_1(1-r^n)}{1-r}$ where $a_1$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. What do we need to use the formula? The number of terms $n$ is ${40}$ because there are ${40}$ numbers from $0$ to $39$. The first term $a_1$ is $8$ because $8\left(-\dfrac78\right)^0 = {8}$. The common ratio $r$ is ${-\dfrac78}$ because it is the base of the exponent in $8\left({-\dfrac78}\right)^k$. Find the sum $(S_n)$ of the series $\begin{aligned} S_n &= \dfrac{a_1(1-r^n)}{1-r} \\\\ S_{{40}}&=\dfrac{{8}(1-\left({-\dfrac78}\right)^{{40}})}{1-\left({-\dfrac78}\right)} \\\\ S_{{{40}}} &\approx 4.25 \end{aligned}$ The answer $ 4.25 $